Why is the smallest normal subgroup containing part of the kernel also in the kernel? In the book "Algebra" by Serge Lang it is stated that, if $f : G \to G'$ is a group homomorphism, $H$ is a subgroup contained in the Kernel of f, and $N$ is the smallest normal subgroup containing $H$, then $N$ is in the Kernel of f.
I would like to fill in the details myself, but somehow I am stuck. How can I use the normality of $N$ to show this ? Does it have to do with the fact that any element in $N$ can be written as a product containing elements of $H$ ? 
Thanks for your help!! 
 A: We can do this by considering the "universal" definition of $N$: $N$ has the following properties:


*

*$H\subseteq N$;

*$N\triangleleft G$; and

*If $K$ is any subgroup such that $H\subseteq K$ and $K\triangleleft G$, then $N\subseteq K$. 


Since $\mathrm{ker}(f)\triangleleft G$, and we are assuming that $H\subseteq \mathrm{ker}(f)$, property 3 implies immediately that $N\subseteq \mathrm{ker}(f)$.
We can also do this by considering the top-down construction of $N$: $N$ is the intersection of all normal subgroups of $G$ that contains $H$:
$$N = \bigcap_{H\subseteq K\triangleleft G} K.$$
Then since $H\subseteq \mathrm{ker}(f)\triangleleft G$, it follows that $N\subseteq\mathrm{ker}(f)$.
Or we can do this by considering the bottom-up construction of $N$; but here you are somewhat incorrect above. You write that "every element of $N$ can be written as a product containing elements of $H$", and this is not quite good enough. The correct statement is that $N$ consists exactly of products of conjugates of elements of $H$:
$$N = \Bigl\{ (g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigm| g_i\in G, h_i\in H, m\in\mathbb{N}\Bigr\}.$$
Given such an element, if we apply $f$ to it we get:
$$\begin{align*}
f\Bigl((g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigr) &= f(g_1)f(h_1)f(g_1)^{-1}\cdots f(g_m)f(h_m)f(g_m)^{-1}\\
&= f(g_1)1f(g_1)^{-1}\cdots f(g_m)1f(g_m)^{-1} \quad\text{(since }H\subseteq \mathrm{ker}(f)\text{)}\\
&= 1,
\end{align*}$$
so for every $n\in N$ we have $f(n)=1$; hence, $N\subseteq \mathrm{ker}(f)$. 
